Answer:
The vertex of this parabola, 
, can be found by completing the square.
Step-by-step explanation:
The goal is to express this parabola in its vertex form:
,
where 
, 
, and 
are constants. Once these three constants were found, it can be concluded that the vertex of this parabola is at 
.
The vertex form can be expanded to obtain:
.
Compare that expression with the given equation of this parabola. The constant term, the coefficient for 
, and the coefficient for 
should all match accordingly. That is:
.
The first equation implies that is equal to 
. Hence, replace the "
" in the second equation with 
to eliminate 
:
.
.
Similarly, replace the "" and the "" in the third equation with and 
, respectively:
.
.
Therefore, 
would be equivalent to 
. The vertex of this parabola would thus be:
.
Use the Product Rule: x^ax^b = x^a + b
6x^3 + 1y^2 + 6
Simplify 3 + 1 to 4
6x^4y^2 + 6
Simplify 2 + 6 to 8
<u>= C) 6x^4 y^8</u>
Answer:
see explanation
Step-by-step explanation:
Assuming you are factoring the expression
Given
4y² + 26y + 30 ← factor out 2 from each term
= 2(2y² + 13y + 15) ← factor the quadratic
Consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term.
product = 2 × 15 = 30 and sum = 13
the factors are 10 and 3
Use these factors to split the y- term
2y² + 10y + 3y + 15 ( factor the first/second and third/fourth terms )
= 2y(y + 5) + 3(y + 5) ← factor out (y + 5) from each term
= (y + 5)(2y + 3)
Thus
4y² + 26y + 30
= 2(y + 5)(2y + 3)
The answer to the question is D
